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Practice Paper 2 Question 5

Find the gradient of the implicitly defined curve \(y^2+2y=x^3+7x\) at all its intersection points with the line \(x=1\).

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  • Hint 1
    Can you differentiate both sides with respect to \(x.\)
  • Hint 2
    How can you rearrange your equation to find \(\frac{dy}{dx}?\)
  • Hint 3
    Try to substitute \(x=1\) into the original equation.


By differentiating both sides with respect to \(x,\) we have \(2y\frac{dy}{dx}+2\frac{dy}{dx}=3x^2+7\) (using the chain rule). Rearranging gives \(\frac{dy}{dx}=\frac{3x^2+7}{2y+2}.\) Now we need to find all intersection points with the line \(x=1.\) These are given by \(y^2+2y-8=0,\) which has solutions \(y\in\{-4,2\}.\) We then plug this into the derivative expression to get the gradient values \(-\frac{5}{3}\) and \(\frac{5}{3}.\)

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